I got a problem wrong on my final exam today. It was in the L'Hospital's rule area. I'd like to understand it before Calculus II next year. I know e^x is always e^x as a derivative, but how do you do x^e? The problem was this:

limit as x-infinity of (e^2x)/x^2e)

Question

I got a problem wrong on my final exam today. It was in the L'Hospital's rule area. I'd like to understand it before Calculus II next year. I know e^x is always e^x as a derivative, but how do you do x^e? The problem was this:

limit as x->infinity of (e^2x)/x^2e)

anonymous · Answer

$$e^x$$ grows faster than any polynomial. no need for l'hopital, limit is infinity

unklerhaukus · Answer

$$d/dx (x^{2e}) = 2ex^{2e-1}$$

anonymous · Answer

limit is infinity.  successive derivatives of 
$$e^{2x}$$ will contain 
$$e^{2x}$$ whereas successive derivatives of the denominator will reduce the power so eventually it will be negative, i.e. the x will be in the numerator.

anonymous · Answer

I wrote that as my justification without l'hospital's, that e^x grows faster, so hopefully I will get partial credit. I guess I didn't think of just writing out 2ex^(2e-1) until you pass the value of e and becomes negative.... that makes sense.